I set the vector V as (a,b,c). I know that you have to multiply out the two vectors, so that is what I did. I then got multiple equations which I used to find the values of a, b, and c. The answer that I got was (2,2,-5), however, that solution only works for the equation on the bottom. I don't know how else to do this problem. Any ideas/answers?
Since $(a,b,c)\times(1,0,-3)=(-3b,3a+c,-b)$, you should solve the system$$\left\{\begin{array}{l}-3b=-6\\3a+c=11\\-b=-2.\end{array}\right.$$You will get that $b=2$ and $c=11-3a$. And, in order to have $(a,2,11-3a).(1,-5,1)=-7$, $a$ shall have to be equal to $4$. So, the answer is $(4,2,-1)$.